To determine the number of turning points, we need to find the number of times the graph changes from increasing to decreasing or vice versa. This occurs at the points where the derivative of the function is equal to zero.
Taking the derivative of g(x):
g'(x) = 3 * (x-5)^2 * (5-x)^4 * [3x^2 - 2x(5+x) + 2(x-5)(5-x)].
We can ignore the constant coefficient and focus on the quantity inside the square brackets, which is a quadratic equation:
3x^2 - 2x(5+x) + 2(x-5)(5-x)
Expanding this equation, we get:
3x^2 - 2x(5+x) + 2(25 - 10x - 5x + x^2)
= 3x^2 - 2x^2 - 10x - 2x^2 + 10x - 50 + 20x - 10x
= -x^2 + 20x - 50
Setting this equal to zero and factoring, we get:
(-x + 10)(x - 5) = 0
So, x = 10 or x = 5.
Now, using the second derivative test, we can determine the nature of these points.
Taking the second derivative of g(x):
g''(x) = 3 * (x-5)^2 * (5-x)^4 * [6x - 2(5+x) + 2(x-5)(-1)]
= 3 * (x-5)^2 * (5-x)^4 * [6x - 10 - 2x -10 + 2x -10]
= 3 * (x-5)^2 * (5-x)^4 * [6x - 30]
Plugging in x = 10:
g''(10) = 3 * (10-5)^2 * (5-10)^4 * [6*10 - 30]
= 3 * 5^2 * (-5)^4 * [60 - 30]
= 3 * 25 * 625 * 30
= 1406250 > 0
Plugging in x = 5:
g''(5) = 3 * (5-5)^2 * (5-5)^4 * [6*5 - 30]
= 0
Since the second derivative is positive at x = 10 and the second derivative is zero at x = 5, we have a local minimum and an inflection point at x = 10.
Therefore, the graph of g(x) has at least one turning point, and the answer is:
1. three at most
Rogelio is asked to sketch a graph of g(x)=3x3(x−5)2(5−x)4 . How many turning points should his sketch include?(1 point)
1. three at most
2. exactly eight
3. eight at most
4. nine at most
1 answer
1 answer